Thanks for the help :) Big O (Definition) We say that g(n) is O(f(n)) - read as,

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Thanks for the help :)

Big O (Definition) We say that g(n) is O(f(n)) - read as, "g of n is Big O of f of n" - when there exists some constants k and n_0 (pronounced as "n nought or as n-sub-zero") such that: g(n) is bound by k * f(n) for problem size n >= n_0. Given the growth function, g(n) = 2n^2 + 4n + 3, determine the Big O. Prove that your analysis is correct by finding a constant k, and a constant no that satisfy the definition of Big O. Given the growth function, g(n) = 4n^3 + n + 1, determine the Big O. Prove that your analysis is correct by finding a constant k, and a constant n_0 that satisfy the definition of Big O. Given the following code segment, determine the Big O. Assume n is the number of elements in the array, myArray. void myMethod(int myArray[], int n) {int num1 = 15; int num2 = 20; int result = num1 + num2; for (int i = 0; i

Explanation / Answer

1) Given that g(n) = 2n^2 + 4n + 3

For this case g(n) = O(n^2)

Let us take k = 3 and n0 = 10

for n = 10

n^2 = 100 and 4n+3 = 43

So n^2 > 4n+3

For n > 10 the difference between n^2 and 4n+3 will only increase. Because n^2 increase at faster pace than 4n+3.

So for n>=10

n^2 > 4n+3 Adding 2n^2 on both sides

3n^2 > 2n^2 + 4n + 3

So 3n^2 > g(n) for n>=10

Hence proved.

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